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Maths question help

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AB is a diameter of a circle centred at the origin O, and P is a point on the circumference of the circle. By considering the position vectors of A, B and P, prove that AP is perpendicular to BP

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If two direction vectors have a dot product of 0, they are perpendicular to each other.

You want to prove AP is perpendicular to BP, lets define AP as P-A and BP as P-B.

Now lets find the dot product of them and it should equal zero after some simplifying. 

AP.BP = (p-a).(p-b)

Expanding the right hand side we get

p^2 - b.p -

Now from the diagram:

Screenshot 5


You can see that b = -a, as they make the diameter. So if we include that in our expression we get

p^2 +

so the a.p cancels out and we're left with

p^2 + a.b

And again we're going to plug in the b = -a

p^2 - a^2

Now p^2 actually just means p.p, and if you dot product a vector by itself, you actually have the square of it's magnitude, i.e. |p|^2

so p^2 - a^2 = |p|^2 - |a|^

And hold on a second.... They are all just points on a circle right? So the modulus of p and a must be equal!

|p|^2 - |a|^2 = 0

There are bits of this proof you may not understand as a lot of it is mainly in the old spec - its not homework from school or anything is it?

Joined: 3 years ago

Posts: 1


Thanks sir, when i attempted it i got it up p•p - a•a and didnt know what to do next. It was a question from exercise 9C in the core pure year 1 book



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